I am interested in understanding what the isometry group of a hemisphere in 4 dimensions is.
Specifically, consider $S^4$ with an isometry $SO(5)$. The $S^4$ is parametrized by four angles: $\theta_{1} \in [0, 2\pi)$ and $\theta_2, \theta_3, \theta_4 \in [0, \pi]$. The line element on a unit $S^4$ is
$ds_{S^4}^2 = d\Omega_4^2 = d\theta_4^2 + \sin^2\theta_4 d\Omega_{3}^2$
Now consider $S^4/\mathbb{Z}_2$ with the $\mathbb{Z}_2$ action on $\theta_4$ restricting it to $[0,\pi/2]$. This yields a hemisphere. What is its isometry group?
Thanks!
$\newcommand{\Reals}{\mathbf{R}}$Generally, let $H_{+} \subset S^{n} \subset \Reals^{n+1}$ be the "upper hemisphere" $\{x_{n+1} \geq 0\}$. Isometries of $H_{+}$ naturally correspond to isometries of $S^{n}$ that preserve the "equator" $\{x_{n+1} = 0\}$. Consequently, the isometry group of $H_{+}$ is isomorphic to $O(n)$, and the identity component of the isometry group is isomorphic to $SO(n)$.